What is modal class? தமிழ் வாழ்க
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The chemical reaction equation for this is
XeF6 + 3H2 ---> Xe + 6HF
Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.
we get n = 0.1433 moles H2
to get the mass of XeF6,
we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.
0.1433 moles H2 x
x 
= 11.71 g XeF6
Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.
XeF6 + 3H2 ---> Xe + 6HF
Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.
we get n = 0.1433 moles H2
to get the mass of XeF6,
we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.
0.1433 moles H2 x
Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.
I hope you understand my working:
1) Finding the mol of NH3 to find the mol of (NH4)2SO4 (ammonium sulfate)
2) Mr of (NH4)2SO4
3) Theoretical yield: The actual grams of (NH4)2SO4 produced when reacting 0.514 mol of NH3 to 0.514 mol H2SO4
4) Using formula of (given grams)/(theoretical grams or actual grams) * 100 = 73%
5) Basic algebra
1) Finding the mol of NH3 to find the mol of (NH4)2SO4 (ammonium sulfate)
2) Mr of (NH4)2SO4
3) Theoretical yield: The actual grams of (NH4)2SO4 produced when reacting 0.514 mol of NH3 to 0.514 mol H2SO4
4) Using formula of (given grams)/(theoretical grams or actual grams) * 100 = 73%
5) Basic algebra